Let $\,\vec v\,$ be a vector with known
analytic form $\,\langle a\,,\,b\rangle\,$:

the horizontal component is $\,a\,$

the vertical component is $\,b\,$

Throughout this section, $\,a\,$ and $\,b\,$ are real numbers.

How can this information be used to find the direction and magnitude of the vector $\,\vec v\,$?
In other words, given the horizontal and vertical components of a vector, what are its magnitude and direction?

As shown in an earlier section, the
magnitude of $\,\vec v = \langle a\,,\,b\rangle\,$ is:
$$
\|\vec v\| = \sqrt{\strut a^2 + b^2}
$$
In words: the magnitude of a vector is the square root of the sum of the squares of
its components.

Magnitude: the Zero Vector

If $\,\vec v = \langle a\,,\,b\rangle\,$ is a vector with $\,\|\vec v\| = 0\,$, then $\,\sqrt{a^2 + b^2} = 0\,$.
The only way this can happen is for both
$\,a\,$ and $\,b\,$ to equal zero.
Thus, $\,\vec v = \langle 0\,,\,0\rangle\,$.
If a vector has magnitude zero, then it must be the zero vector.

‘Standard Direction’ Angle Conventions

Place the tail of the vector
at the origin of a rectangular coordinate system.

Indicate the vector's direction using an angle $\,\theta\,$ laid off
as follows:

start at the positive $x$-axis

positive angles are swept out in a counterclockwise direction(start by going up)

negative angles are swept out in a clockwise direction(start by going down)

When it's necessary to emphasize that this direction scheme is being used,
it will be referred to as the standard direction.

Standard direction is not unique. For example, $\,\theta = 30^\circ\,$
and $\,\theta = -330^\circ\,$ both describe a vector pointing the same way.
In what follows, the symbol $\,\theta\,$ (theta) always refers to standard direction, and
simple values of $\,\theta\,$ are used.

standard direction:
a positive angle $\,\theta$

standard direction:
a negative angle $\,\theta$

Standard Direction for Vectors pointing Up/Down/Left/Right

When at least one component (horizontal or vertical) of a vector equals zero,
then no formula is needed to find standard direction.
Here's a table that summarizes the results for $\,\vec v = \langle a\,,\, b\rangle\,$:

horizontal/vertical components

arrow representation

standard direction, $\,\theta$

the zero vector:
$\,a = 0\,$ and $\,b = 0\,$

(no arrow representation)

the zero vector has no direction;
the direction of $\,\langle 0,0\rangle\,$ is undefined

$\,\vec v = \langle a\,,\,0\rangle\,$ with $\,a > 0\,$

vector points to the right

$\,\theta = 0^\circ\,$

$\,\vec v = \langle a\,,\,0\rangle\,$ with $\,a < 0\,$

vector points to the left

$\,\theta = 180^\circ\,$

$\,\vec v = \langle 0\,,\,b\rangle\,$ with $\,b > 0\,$

vector points up

$\,\theta = 90^\circ\,$

$\,\vec v = \langle 0\,,\,b\rangle\,$ with $\,b < 0\,$

vector points down

$\,\theta = 270^\circ\,$ or $\,\theta = -90^\circ\,$

Formula for Standard Direction for Vectors in Quadrants I and IV

For an arbitrary vector, the arctangent function gives convenient formulas for
standard direction. However, different formulas are needed in
different quadrants.

Why are different formulas needed?
The arctangent only gives angles between $\,-90^\circ\,$
and $\,90^\circ\,$, so (without adjustment) it can only cover vectors
in quadrants IV and I. Keep reading!

Things are similarly easy in quadrant IV. Here, $\,\frac ba\,$ is negative,
and $\,\arctan(\frac ba)\,$ returns an angle between $\,-90^\circ\,$ and
$\,0^\circ\,$. Again, this angle provides a correct direction for the vector.
If you want a positive angle for the direction (instead of the negative
one provided by the arctangent), you can use the formula
$\,360^\circ + \arctan(\frac ba) \,$.

The formula for the (standard) direction
of a vector $\,\vec v = \langle a\,,\,b\rangle\,$
depends upon the signs of $\,a\,$ and $\,b\,$, as indicated below:

The ‘Reference Angle’ Method for finding (standard) Direction

Instead of using the formulas for direction discussed above, you can always find the reference angle for the vector, and then adjust.
The reference angle is the acute angle between the vector and the $x$-axis.

EXAMPLE

Find the standard direction of $\,\vec v = \langle -2\,,\,-5\rangle\,$.
Report as a positive
angle in the interval $\,[0^\circ,360^\circ)\,$.